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I tried to find a closed-form expression for the integral

$$\int\frac{1}{a+b\sec(x)}\>dx$$ and, afterwards, set $a=0$, $b=1$ to recover the result for the simplified integral $$\int\cos(x)\>dx=\sin(x)+C$$

The above integral was inspired by Américo Tavares’s rather illuminating solution to Ways to evaluate $\sec(t)$,

Here's the progress I've made which makes use of the integral

$$(\star)\int\frac{dx}{m^2+n^2x^2}=\frac{1}{mn}\arctan\left(\frac{nx}{m}\right)\quad m,n>0$$

Using $\cos(x)=2\cos^2(x/2)-1$, we get the following

$$I=\int\frac{dx}{a+b\sec(x)}=\int\frac{\cos(x)}{a\cos(x)+b}dx=\int\frac{2\cos^2(x/2)-1}{2a\cos^2(x/2)+b-a}dx\\=2\underbrace{\int\frac{dx}{2a+(b-a)\sec^2(x/2)}}_\color\red{(1)}-\underbrace{\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx}_\color\red{(2)}$$

$\color\red{(2)}$ becomes

$$\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx=\int\frac{\sec^2(x/2)}{a+b+(b-a)\tan^2(x/2)}dx\\=\frac{2}{\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)$$

from the substitution $u=\tan(x/2)$ and then applying $(\star)$.

For tackling $\color\red{(1)}$, the same u-substitution and applying partial fractions converts the integral into

$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=2\int\frac{du}{(a+b+(b-a)u^2)(1+u^2)}\\=2\left[\frac{a-b}{2a}\int\frac{du}{(a+b+(b-a)u^2)}+\frac{1}{2a}\int\frac{du}{1+u^2}\right]$$

Thus

$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}u\right)+\frac{1}{a}\arctan(u)\\=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)+\frac{1}{a}\arctan(\tan(x/2)))$$

However, the problem is that setting $a=0$ afterwards results in the denominator becoming zero. Also, I couldn't think another approach to this.

Any help?

Quanto
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Anson Pang
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2 Answers2

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As @Hussain-Alqatari commented, using immediately the tangent half angle substitution, you have $$I=\int \frac{dx}{a+b\sec(x)}=\int \frac{2(1-t^2)}{ (b-a)t^4+2 b t^2+(a+b)}\,dt$$ Assuming $b\neq a$ $$I=\frac 2{b-a}\int \frac {1-t^2}{(t^2+1)(t^2-\alpha)}\,dt \qquad \text{with} \qquad \alpha=\frac{a+b}{a-b}$$ Using partial fraction decomposition $$ \frac {1-t^2}{(t^2+1)(t^2-\alpha)}=-\frac{2}{1+\alpha }\frac 1{t^2+1}+\frac{1-\alpha }{1+\alpha }\frac 1 {t^2-\alpha}$$ which is very simple $$\color{blue}{I=\frac{2 }{a}\tan ^{-1}(t)-\frac {2b}{a \sqrt{a^2-b^2}}\tanh ^{-1} \left(\sqrt{\frac{a-b}{a+b}}t\right)}$$

Claude Leibovici
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    But $(a,b)=(0,1)$ still gives a zero denominator and complex radicals. Is there a closed form that circumvents this? – Anson Pang Oct 03 '21 at 14:39
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To derive the result valid for $a\to 0$, integrate as follows

\begin{align} \int \frac{1}{a+b\sec x}dx =& \frac1a \int \left(1- \frac{b}{b+a\cos x}\right)dx\\ =& \frac1a \bigg(x-2b\int \frac{d(\tan\frac x2)}{(b+a)+(b-a)\tan^2\frac x2}\bigg)\\ =& \frac1a \bigg( x- \frac{2b}{\sqrt{b^2-a^2}}\tan^{-1}\frac{\tan\frac x2}{\sqrt{\frac{b+a}{b-a}} \>}\bigg) +C\\ \end{align} Note that $$\lim_{a\to 0 } \frac{2b}{\sqrt{b^2-a^2}}\tan^{-1}\frac{\tan\frac x2}{\sqrt{\frac{b+a}{b-a}} } = x - \frac ab \sin x+O(a^2) $$ Thus $$ \lim_{a\to 0}\> \int \frac{1}{a+b\sec x}dx=\frac1b \sin x+C $$ as expected.

Quanto
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